If on one of the sides of a triangle, from its
extremities, there be constructed two straight lines
meeting within the triangle, the straight lines so
constructed will be less than the remaining two sides
of the triangle, but will contain a greater angle.

On BC, one of the sides of
the triangle ABC, from
its extremities, B,
C, let the two straight lines
BD, DC
be constructed meeting within the triangle;
I say that BD,
DC are less than the remaining two
sides of the triangle BA,
AC, but contain an
angle BDC greater than the
angle BAC.

For let BD be drawn through
to E.

ABCDE

Then, since in any triangle two sides are greater than
the remaining one,
[I. 20]
therefore, in the triangle ABE,
the two sides
AB, AE
are greater than BE.
Let EC be added to each,
therefore BA,
AC are greater than
BE, EC.

Again, since, in the
triangle CED,
the two sides CE,
ED are greater
than CD,
let DB be added to each;
therefore CE,
EB are greater
than CD, DB.

But BA, AC,
were proved greater than
BE,
EC;
therefore BA,
AC are
much greater than BD,
DC.

Again, since in any triangle the exterior angle is
greater than the interior and opposite angle,
[I. 16]
therefore in the
triangle CDE,
the exterior angle BDC is
greater than the angle CED.

For the same reason, moreover, in the
triangle ABE also,
the exterior angle CEB is greater
than the angle BAC.
But the angle BDC was proved greater than
the angle CEB;
therefore the angle BDC is
much greater than the angle BAC.